Unlocking the Secrets of Solving Quadratic Equations Graphically

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Finding Solutions of Quadratic Equations

Consider a quadratic function in standard form.

f (x) = \frac{3}{4} x^{2} - 3 x

Recall the steps followed to graph a quadratic function in standard form. Use the applet to draw the graph.

Now, consider the quadratic equation

f (x) = 0 .

f (x) = 0 \Leftrightarrow \frac{3}{4} x^{2} - 3 x = 0

Think about the definitions of an

x -

intercept and a solution of an equation. How do the

x -

intercepts of the quadratic function and the solutions of the quadratic equation relate to each other?

Discussion

Quadratic Equation

A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is $2 .$ A quadratic equation can be written in standard form as follows.

$a x^{2} + b x + c = 0$

Here, the leading coefficient

a

is non-zero, which guarantees that the

x^{2} -

term is present. Solving a quadratic equation means finding the zeros of the related quadratic function. Therefore, quadratic equations can have at most two solutions.

Quadratic Equation a x^{2} + b x + c = 0 Related Function y = a x^{2} + b x + c

This type of equation can be solved using several methods, such as graphing and factoring.

Concept

Number of Solutions of a Quadratic Equation

The solutions of a quadratic equation can be interpreted graphically as the zeros of the related quadratic function. Therefore, the number of solutions of a quadratic equation is the same as the number of zeros of the related function.

Quadratic Equation a x^{2} + b x + c = 0 Quadratic Function y = a x^{2} + b x + c

If the function has two zeros, the equation

a x^{2} + b x + c = 0

has two solutions. Similarly, if the function has one zero, the equation has one solution. Finally, if the function does not have any zeros, the equation has no real solutions.

Pop Quiz

Using a Graph to Find the Number of Solutions of a Quadratic Equation

For each quadratic equation, draw its related quadratic function to determine the number of solutions.

Finding number of solutions of quadratic equations

Example

Finding the Time When a Rocket Reaches a Certain Height

Paulina models the flight of a rocket she built using a quadratic function, where $h$ is the height of the rocket in meters after $t$ seconds.

a After how many seconds does the rocket reach a height of

192

meters?

b After how many seconds does the rocket reach a height of

196

meters?

Hint

a Set the quadratic function equal to

192 .

Then, rewrite it in standard form to solve by graphing.

b Set the quadratic function equal to

196 .

Then, rewrite it in standard form to solve by graphing.

Solution

a To find after how many seconds the rocket is at a height of

192

meters,

192

will be substituted for

h

into the given quadratic function.

h = - 16 t^{2} + 112 t ⇓ 192 = - 16 t^{2} + 112 t

This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.

192 = - 16 t^{2} + 112 t

SubEqn

$LHS - 192 = RHS - 192$

0 = - 16 t^{2} + 112 t - 192

RearrangeEqn

Rearrange equation

- 16 t^{2} + 112 t - 192 = 0

Now the function related to the equation can be written.

Equation - 16 t^{2} + 112 t - 192 = 0 ⇓ Related Function f (t) = - 16 t^{2} + 112 t - 192

The solutions to the equation are the

x -

intercepts of the function's graph. To graph the related quadratic function, its characteristics must be first determined.

$f (t) = - 16 t^{2} + 112 t - 192$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 4)$	$x = 3.5$	$(0, - 192)$

In addition to the vertex and the $y -$ intercept, the reflection of the $y -$ intercept across the axis of symmetry, which is at $(7, - 192),$ is also on the parabola. With this information, the graph of the function can be drawn.

The parabola intersects the

x -

axis twice. The points of intersection are

(3, 0)

and

(4, 0) .

Therefore, the equation

- 16 t^{2} + 112 t - 192 = 0

has two solutions,

t_{1} = 3

and

t_{2} = 4 .

- 16 t^{2} + 112 t - 192 = 0 ↙ ↘ t_{1} = 3 t_{2} = 4

These solutions mean that the rocket is at a height of

192

meters after

3

and after

4

seconds.

b Following the same steps as above, the time at which the rocket is at

196

meters can be calculated.

h = - 16 t^{2} + 112 t ⇓ 196 = - 16 t^{2} + 112 t

This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.

196 = - 16 t^{2} + 112 t

SubEqn

$LHS - 196 = RHS - 196$

0 = - 16 t^{2} + 112 t - 196

RearrangeEqn

Rearrange equation

- 16 t^{2} + 112 t - 196 = 0

Now the function related to the equation can be written.

Equation - 16 t^{2} + 112 t - 196 = 0 ⇓ Related Function g (t) = - 16 t^{2} + 112 t - 196

Again, the solutions to the equation are the

x -

intercepts of the parabola. Identify the characteristics of the function to be able to draw it.

$g (t) = - 16 t^{2} + 112 t - 196$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 0)$	$x = 3.5$	$(0, - 196)$

The reflection of the $y -$ intercept across the axis of symmetry is at $(7, - 196),$ which is also on the parabola. Now, the graph can be drawn.

This time, there is only one $x -$ intercept.

The point of intersection is

(3.5, 0) .

Therefore, the equation

- 16 t^{2} + 112 t - 196 = 0

has one solution,

t = 3.5 .

- 16 t^{2} + 112 t - 196 = 0 ↓ t = 3.5

This solution means that the rocket is at a height of

196

meters after

3.5

seconds.

Example

Modeling Heights of Arrows With Quadratic Equations

Tiffaniqua and Ramsha are good at archery. They want to determine whether the arrows they shoot will collide in the air or not. Tiffaniqua takes her shot from a tree and Ramsha takes her shot from the ground just below Tiffaniqua. They wrote two quadratic functions to model the heights of the arrows in meters.

Tiffaniqua : Ramsha : y = - 1.2 x^{2} + 4.8 x + 8 y = - 0.5 (x - 4)^{2} + 8

In these equations,

x

represents the horizontal distance in meters from the point where the arrow is shot.

a Write a quadratic equation in standard form that will help determine whether the arrows will collide.

b Find the answer that fits the context of the situation by solving the equation graphically.

c Suppose a coordinate plane is placed on the diagram so that Ramsha is at the origin. How far is Ramsha from the point where the arrows collide? Write the answer in exact form.

Hint

a At the collision point of the arrows, the functions have the same height. Therefore, equate the given quadratic functions.

b The equation written in the previous part has a related function. Draw the graph of this function.

c Find the coordinates of the point where the arrows collide. Then, use the Distance Formula.

Solution

a At the collision point of the arrows, if there is any, the functions have the same

y -

value. Therefore, to find this point, the right-hand sides of the equations can be equated.

{y = - 1.2 x^{2} + 4.8 x + 8 y = - 0.5 (x - 4)^{2} + 8 ⇓ - 1.2 x^{2} + 4.8 x + 8 = - 0.5 (x - 4)^{2} + 8

This resulting quadratic equation needs to be written in standard form.

- 1.2 x^{2} + 4.8 x + 8 = - 0.5 (x - 4)^{2} + 8

Simplify right-hand side

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

- 1.2 x^{2} + 4.8 x + 8 = - 0.5 (x^{2} - 8 x + 16) + 8

Distr

Distribute $- 0.5$

- 1.2 x^{2} + 4.8 x + 8 = - 0.5 x^{2} + 4 x - 8 + 8

SubTerm

Subtract term

- 1.2 x^{2} + 4.8 x + 8 = - 0.5 x^{2} + 4 x

SubEqn

$LHS - 4 x = RHS - 4 x$

- 1.2 x^{2} + 0.8 x + 8 = - 0.5 x^{2}

AddEqn

$LHS + 0.5 x^{2} = RHS + 0.5 x^{2}$

- 0.7 x^{2} + 0.8 x + 8 = 0

The solutions to this quadratic equation will give the horizontal distance to the collision point.

b The quadratic equation found in Part A will solved graphically. To do so, its related function will be drawn first.

Quadratic Equation - 0.7 x^{2} + 0.8 x + 8 = 0 ⇓ Related Function y = - 0.7 x^{2} + 0.8 x + 8

To be able to graph the function, some of its characteristics should be determined.

$y = - 0.7 x^{2} + 0.8 x + 8$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(\frac{4}{7}, \frac{2 8 8}{3 5})$	$x = \frac{4}{7}$	$(0, 8)$

The reflection of the

y -

intercept across the axis of symmetry

(\frac{8}{7}, 8)

is also on the parabola. Since these points are very close to each other, another point can be plotted by randomly choosing an

x -

value and calculating its corresponding

y -

value. Try

x = - 3 .

y = - 0.7 x^{2} + 0.8 x + 8

Substitute

$x = - 3$

y = - 0.7 (- 3)^{2} + 0.8 (- 3) + 8

Evaluate right-hand side

CalcPow

Calculate power

y = - 0.7 (9) + 0.8 (- 3) + 8

Multiply

y = - 6.3 - 2.4 + 8

AddSubTerms

Add and subtract terms

y = - 0.7

Therefore, the point

(- 3, - 0.7)

lies on the curve. The reflection of this point across the axis of symmetry is

(\frac{2 9}{7}, - 0.7),

which also lies on the parabola. With this information, the graph can be drawn.

The $x -$ intercepts of the graph can now be identified.

The parabola intersects the

x -

axis twice. From the graph only one of the intercepts is identified easily, which is

(4, 0) .

The other intercept is between

- 3

and

- 2 .

However, this intercept can be ignored because

x

represents horizontal distance and distance cannot be negative. Therefore, in this context, only

x = 4

makes sense.

Equation : Solution : - 0.7 x^{2} + 0.8 x + 8 = 0 x = 4

This solution means that the horizontal distance from where Tiffaniqua and Ramsha stand to the point of collision is

4

meters.

c Place a coordinate plane so that Ramsha is at the origin. In Part B, the horizontal distance to the point where the arrows collide was found to be

4

meters

To find how far Ramsha is to the collision point, the Distance Formula will be used.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

To do so, the

y -

coordinate of the collision point needs to be found. To accomplish this, substitute

4

into the either of the quadratic functions, and evaluate.

y = 0.5 (x - 4)^{2} + 8

Substitute

$x = 4$

y = 0.5 (4 - 4)^{2} + 8

Evaluate right-hand side

DiffZero

$a - a = 0$

y = 0.5 (0)^{2} + 8

CalcPow

Calculate power

y = 0.5 (0) + 8

ZeroPropMult

Zero Property of Multiplication

y = 0 + 8

IdPropAdd

\IdPropAdd

y = 8

Therefore, the arrows collide at the point

(4, 8) .

Now the distance from Ramsha to the point where the arrows collide can be found.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(0, 0)$ & $(4, 8)$

d = (4 - 0)^{2} + (8 - 0)^{2}

Evaluate right-hand side

SubTerms

Subtract terms

d = 4^{2} + 8^{2}

CalcPow

Calculate power

d = 16 + 64

AddTerms

Add terms

d = 80

SplitIntoFactors

Split into factors

d = 16 \cdot 5

SqrtProd

$a \cdot b = a \cdot b$

d = 16 \cdot 5

CalcRoot

Calculate root

d = 4 \cdot 5

Multiply

d = 45

Ramsha is

45

meters away from the collision point of the arrows.

Pop Quiz

Finding Solutions of Quadratic Equations

Find the solutions to each quadratic equation by graphing its related function. If there are two solutions, write the smaller solution first.

Closure

Graphing Each Side of an Equation

In this lesson, real life situations that can be modeled by quadratic equations were solved graphically. Solving a quadratic equation graphically can be done in three steps.

Step I : Step II : Step III : Write the Equation in Standard Form Graph the Related Function Find the x - intercepts

In general, when solving equations of form

f (x) = g (x),

the equation is rearranged and the graph of

y = f (x) - g (x)

is drawn. The zeros of the graph are solutions to the equation.

Another way to graphically solve an equation is to graph

y = f (x)

and

y = g (x)

separately, then look for the points of intersection.

As can be seen, both methods give the same solutions to the given examples. Therefore, both methods can be used to solve an equation graphically.

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Hint

Solution

Hint

Solution